Question

A parallel plate capacitor having rectangular
plates of area of cross-section 6.45 x 10^-4 m² each is kept in free space. The separation
between the plates is 1.5 x 10^-3 m and a
voltage of 15 V is applied across these plates. Calculate the capacitance of the capacitor. If a material of dielectric constant 6.0 is introduced in the region between the two plates, calculate the capacitance and the
electric displacement D .

Answer 1

The capacitance of a parrallel-plate capacitor with known dimensions can be obtained by means of the following expression:

$$C_0 = \frac{\varepsilon_0 S}{d}$$

Substituting the numerical values, we obtain:

$$C_0 = \frac{8.85 \cdot 10^{-12} \cdot 6.45 \cdot 10^{-4}}{1.5 \cdot 10^{-3}} \approx 3.8 \cdot 10^{-12} \, F = 3.8 \, pF$$

When the dielectric material is introduced between the plates, the capacitance is increased by the factor of dielectric constant value, i.e.,

$$C_{\varepsilon} = \varepsilon C_0$$

Substituting the numerical values, we obtain:

$$C_{\varepsilon} = 6 \cdot 3.8 \cdot 10^{-12} \approx 2.3 \cdot 10^{-11} \, F = 23 \, pF$$

The electric displacement is defined through the electric field value in the way as follows:

$$D = \varepsilon_0 \varepsilon E$$

where

$$E = \frac{V}{d}$$

Finally, we obtain:

$$D = \frac{\varepsilon_0 \varepsilon V}{d}$$

Substituting the numerical values, we obtain:

$$D = \frac{8.85 \cdot 10^{-12} \cdot 6 \cdot 15}{1.5 \cdot 10^{-3}} \approx 5.3 \cdot 10^{-7} \, C/m^2 = 0.53 \, \mu C/m^2$$

Answer: 3.8 pF; 23 pF, 0.53 muC/m²